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#python 3.7.x ile derlenmiştir. | |
# [2178, 21978, 219978] | |
def is_palindrom(foo,bar): | |
n1=list(str(foo)) | |
n2=list(str(bar)) | |
if n1==n2[::-1]: | |
return True | |
else: | |
return False |
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from bs4 import BeautifulSoup | |
import time | |
import urllib.request as urlac | |
while True: | |
url = "https://www.doviz.com/bitcoin" | |
url_oku = urlac.urlopen(url) | |
soup = BeautifulSoup(url_oku, 'html.parser') | |
vericek = soup.find_all('span',attrs={'class':'color-red'}) |
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#Pi sayısının İlk yüz basamağı; 3.14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 | |
#74944 59230 78164 06286 20899 86280 34825 34211 70679…. gibidir. Virgülden sonraki ilk 100 basamak | |
#içerisinde ard arda gelen 6 basamağın çarpımlarının maksimumu ve minimumu arasındaki fark kaçtır. | |
i="1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679" | |
carpim=1 | |
dizi=[] | |
p=True | |
x=0 |
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i=1923 | |
x=0 | |
c=[] | |
while(i<=2017): | |
if(i%4==0): | |
x=x+366 | |
else: | |
x=x+365 | |
if(x%7 == 0 or x%7==6): | |
c.append(i) |
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class republic(): | |
def __init__(self,year,day): | |
self.year=year | |
self.day=day | |
i=1923 | |
x=0 | |
c=[] | |
while(i<=2017): |
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using System; | |
using System.Collections.Generic; | |
using System.Linq; | |
using System.Text; | |
using System.Data; | |
using System.Data.OleDb; | |
namespace WindowsFormsApplication1 | |
{ | |
class scvp |
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using System; | |
using System.Collections.Generic; | |
using System.ComponentModel; | |
using System.Data; | |
using System.Drawing; | |
using System.Linq; | |
using System.Text; | |
using System.Windows.Forms; | |
using System.Data; | |
using System.Data.OleDb; |
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# Özgür KARAÇAM 14060824 | |
# 20.01.2016 11.34 | |
#Soru 1: (30 p) Özyineleme | |
#a) Fonksiyonun geri döndürme yeteneği yokken ve döngü kullanmaksızın 10'luk tabandaki | |
#bir sayıyı herhangi bir tabana dönüştüren baseConvert işlevi yazınız. İpucu: özyineleme ve | |
#stack yardımcı olabilir. Örnek: baseConvert(123, 7) 123 sayısının 7 taban karşılığı üretilecek. | |
a="0123456789ABCDEF" | |
s=Stack() | |
def cevir(sayi,taban): |
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#SORU-4: 196 Problemi | |
#Kullanıcıdan alınan en az iki basamaklı sayı (örneğin abc) alıp, bu sayının ters yazılmışını (cba) hesaplayan. | |
#Ve bu sayıların toplamı palindrom (düz ve ters yazılışı aynı olan. Örneğin 12321) olmasına kadar hesaba | |
#devam eden programı yazınız. Bu kuralla hesaplanabilen en küçük sayı 196 olduğundan bu probleme 196 problemi | |
#denilir. Örnek, | |
#Girdi: 57 | |
#Ara hesaplar | |
# 57 + 75 = 132 | |
#132 + 231 = 363 (palindrom, programı sonlandır) |
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class Stack: | |
def __init__(self): | |
self.alist=[] | |
def pop(self): | |
return self.alist.pop() | |
def push(self,a): | |
self.alist.append(a) | |
def show(self): | |
return self.alist | |
def peek(self): |
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